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Thread: squares

  1. #1
    Member Veronica1999's Avatar
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    Jupimar
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    squares

    I can't seem to figure out the logic behind this problem.
    Would this problem have something to do with mode?
    I know I am missing the main point of this problem.
    Attached Thumbnails Attached Thumbnails squares-squares.jpg  
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  2. #2
    Senior Member
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    Crna Gora
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    Re: squares

    Since $\displaystyle 2^{n-12} +49 $ is an odd number it must be perfect square also . So:

    $\displaystyle x^2=2^{n-12}+49 \Rightarrow 2^{n-12}=x^2-49$

    Now, if we make substitution $\displaystyle y=n-12$ we can write :

    $\displaystyle 2^y=x^2-49 \Rightarrow y= \log_{2} (x^2-49)$

    This equality has integer solutions only if $\displaystyle x^2-49$ is power of $\displaystyle 2$ so $\displaystyle x$ must be an odd number so we shall make substitution $\displaystyle x=2m-1$ and therefore :

    $\displaystyle y=\log_{2} (4m^2-4m-48)=\log_{2}(4 \cdot(m^2-m-12))$

    $\displaystyle =2+\log_{2} (m^2-4m+3m-12)=$

    $\displaystyle =2+\log_{2} (m+3)(m-4)=2+\log_{2} (m+3)+\log_{2}(m-4)$

    It is obvious that both $\displaystyle m+3$ and $\displaystyle m-4$ must be powers of $\displaystyle 2$ and this is possible only if $\displaystyle m=5$ , therefore :

    $\displaystyle y=2+3+0 \Rightarrow y=5 \Rightarrow n-12=5 \Rightarrow n=17 .$
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